2D masīvu var definēt kā masīvu masīvu. 2D masīvs ir sakārtots kā matricas, kuras var attēlot kā rindu un kolonnu kolekciju.
Tomēr 2D masīvi tiek izveidoti, lai ieviestu relāciju datu bāzes datu struktūru. Tas nodrošina ērtu datu masveida glabāšanu vienlaikus, ko var nodot jebkuram skaitam funkciju, kur vien tas ir nepieciešams.
Kā deklarēt 2D masīvu
Divdimensiju masīva deklarēšanas sintakse ir ļoti līdzīga viendimensiju masīva sintakse, kas norādīta šādi.
pārvērst par virkni java
int arr[max_rows][max_columns];
tomēr tas rada datu struktūru, kas izskatās šādi.
Augšējā attēlā redzams divdimensiju masīvs, elementi ir sakārtoti rindu un kolonnu veidā. Pirmās rindas pirmo elementu attēlo a[0][0], kur pirmajā rādītājā parādītais skaitlis ir šīs rindas numurs, savukārt otrajā rādītājā parādītais skaitlis ir kolonnas numurs.
Kā piekļūt datiem 2D masīvā
Sakarā ar to, ka 2D masīvu elementiem var piekļūt nejauši. Līdzīgi kā viendimensiju masīviem, mēs varam piekļūt atsevišķām šūnām 2D masīvā, izmantojot šūnu indeksus. Konkrētai šūnai ir pievienoti divi indeksi, viens ir tās rindas numurs, bet otrs ir kolonnas numurs.
Tomēr mēs varam saglabāt vērtību, kas saglabāta jebkurā konkrētā 2D masīva šūnā, uz kādu mainīgo x, izmantojot šādu sintaksi.
int x = a[i][j];
kur i un j ir attiecīgi šūnas rindas un kolonnas numurs.
Būla valoda c
Mēs varam piešķirt katrai 2D masīva šūnai 0, izmantojot šādu kodu:
for ( int i=0; i<n ;i++) { for (int j="0;" j<n; j++) a[i][j]="0;" } < pre> <h2>Initializing 2D Arrays </h2> <p>We know that, when we declare and initialize one dimensional array in C programming simultaneously, we don't need to specify the size of the array. However this will not work with 2D arrays. We will have to define at least the second dimension of the array. </p> <p>The syntax to declare and initialize the 2D array is given as follows. </p> <pre> int arr[2][2] = {0,1,2,3}; </pre> <p>The number of elements that can be present in a 2D array will always be equal to ( <strong>number of rows * number of columns</strong> ). </p> <p> <strong>Example :</strong> Storing User's data into a 2D array and printing it. </p> <p> <strong>C Example : </strong> </p> <pre> #include void main () { int arr[3][3],i,j; for (i=0;i<3;i++) { for (j="0;j<3;j++)" printf('enter a[%d][%d]: ',i,j); scanf('%d',&arr[i][j]); } printf('
printing the elements ....
'); for(i="0;i<3;i++)" printf('
'); printf('%d ',arr[i][j]); < pre> <h3>Java Example</h3> <pre> import java.util.Scanner; publicclass TwoDArray { publicstaticvoid main(String[] args) { int[][] arr = newint[3][3]; Scanner sc = new Scanner(System.in); for (inti =0;i<3;i++) { for(intj="0;j<3;j++)" system.out.print('enter element'); arr[i][j]="sc.nextInt();" system.out.println(); } system.out.println('printing elements...'); for(inti="0;i<3;i++)" system.out.print(arr[i][j]+' '); < pre> <h3>C# Example </h3> <pre> using System; public class Program { public static void Main() { int[,] arr = new int[3,3]; for (int i=0;i<3;i++) { for (int j="0;j<3;j++)" console.writeline('enter element'); arr[i,j]="Convert.ToInt32(Console.ReadLine());" } console.writeline('printing elements...'); i="0;i<3;i++)" console.writeline(); console.write(arr[i,j]+' '); < pre> <h2>Mapping 2D array to 1D array </h2> <p>When it comes to map a 2 dimensional array, most of us might think that why this mapping is required. However, 2 D arrays exists from the user point of view. 2D arrays are created to implement a relational database table lookalike data structure, in computer memory, the storage technique for 2D array is similar to that of an one dimensional array. </p> <p>The size of a two dimensional array is equal to the multiplication of number of rows and the number of columns present in the array. We do need to map two dimensional array to the one dimensional array in order to store them in the memory.</p> <p>A 3 X 3 two dimensional array is shown in the following image. However, this array needs to be mapped to a one dimensional array in order to store it into the memory. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-2.webp" alt="DS 2D Array"> <br> <p>There are two main techniques of storing 2D array elements into memory </p> <h3>1. Row Major ordering </h3> <p>In row major ordering, all the rows of the 2D array are stored into the memory contiguously. Considering the array shown in the above image, its memory allocation according to row major order is shown as follows. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-3.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> row of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last row.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-4.webp" alt="DS 2D Array"> <br> <h3>2. Column Major ordering </h3> <p>According to the column major ordering, all the columns of the 2D array are stored into the memory contiguously. The memory allocation of the array which is shown in in the above image is given as follows.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-5.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> column of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last column of the array. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-6.webp" alt="DS 2D Array"> <br> <h2>Calculating the Address of the random element of a 2D array </h2> <p>Due to the fact that, there are two different techniques of storing the two dimensional array into the memory, there are two different formulas to calculate the address of a random element of the 2D array. </p> <h3>By Row Major Order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = B. A. + (i * n + j) * size </pre> <p>where, B. A. is the base address or the address of the first element of the array a[0][0] . </p> <p> <strong>Example : </strong> </p> <pre> a[10...30, 55...75], base address of the array (BA) = 0, size of an element = 4 bytes . Find the location of a[15][68]. Address(a[15][68]) = 0 + ((15 - 10) x (68 - 55 + 1) + (68 - 55)) x 4 = (5 x 14 + 13) x 4 = 83 x 4 = 332 answer </pre> <h3>By Column major order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = ((j*m)+i)*Size + BA </pre> <p>where BA is the base address of the array. </p> <p> <strong>Example:</strong> </p> <pre> A [-5 ... +20][20 ... 70], BA = 1020, Size of element = 8 bytes. Find the location of a[0][30]. Address [A[0][30]) = ((30-20) x 24 + 5) x 8 + 1020 = 245 x 8 + 1020 = 2980 bytes </pre> <hr></3;i++)></pre></3;i++)></pre></3;i++)></pre></n> Elementu skaits, kas var būt 2D masīvā, vienmēr būs vienāds ar ( rindu skaits * kolonnu skaits ).
Piemērs : Lietotāja datu glabāšana 2D masīvā un drukāšana.
C Piemērs:
#include void main () { int arr[3][3],i,j; for (i=0;i<3;i++) { for (j="0;j<3;j++)" printf(\'enter a[%d][%d]: \',i,j); scanf(\'%d\',&arr[i][j]); } printf(\'
printing the elements ....
\'); for(i="0;i<3;i++)" printf(\'
\'); printf(\'%d \',arr[i][j]); < pre> <h3>Java Example</h3> <pre> import java.util.Scanner; publicclass TwoDArray { publicstaticvoid main(String[] args) { int[][] arr = newint[3][3]; Scanner sc = new Scanner(System.in); for (inti =0;i<3;i++) { for(intj="0;j<3;j++)" system.out.print(\'enter element\'); arr[i][j]="sc.nextInt();" system.out.println(); } system.out.println(\'printing elements...\'); for(inti="0;i<3;i++)" system.out.print(arr[i][j]+\' \'); < pre> <h3>C# Example </h3> <pre> using System; public class Program { public static void Main() { int[,] arr = new int[3,3]; for (int i=0;i<3;i++) { for (int j="0;j<3;j++)" console.writeline(\'enter element\'); arr[i,j]="Convert.ToInt32(Console.ReadLine());" } console.writeline(\'printing elements...\'); i="0;i<3;i++)" console.writeline(); console.write(arr[i,j]+\' \'); < pre> <h2>Mapping 2D array to 1D array </h2> <p>When it comes to map a 2 dimensional array, most of us might think that why this mapping is required. However, 2 D arrays exists from the user point of view. 2D arrays are created to implement a relational database table lookalike data structure, in computer memory, the storage technique for 2D array is similar to that of an one dimensional array. </p> <p>The size of a two dimensional array is equal to the multiplication of number of rows and the number of columns present in the array. We do need to map two dimensional array to the one dimensional array in order to store them in the memory.</p> <p>A 3 X 3 two dimensional array is shown in the following image. However, this array needs to be mapped to a one dimensional array in order to store it into the memory. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-2.webp" alt="DS 2D Array"> <br> <p>There are two main techniques of storing 2D array elements into memory </p> <h3>1. Row Major ordering </h3> <p>In row major ordering, all the rows of the 2D array are stored into the memory contiguously. Considering the array shown in the above image, its memory allocation according to row major order is shown as follows. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-3.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> row of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last row.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-4.webp" alt="DS 2D Array"> <br> <h3>2. Column Major ordering </h3> <p>According to the column major ordering, all the columns of the 2D array are stored into the memory contiguously. The memory allocation of the array which is shown in in the above image is given as follows.</p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-5.webp" alt="DS 2D Array"> <br> <p>first, the 1<sup>st</sup> column of the array is stored into the memory completely, then the 2<sup>nd</sup> row of the array is stored into the memory completely and so on till the last column of the array. </p> <br> <img src="//techcodeview.com/img/ds-tutorial/80/2d-array-6.webp" alt="DS 2D Array"> <br> <h2>Calculating the Address of the random element of a 2D array </h2> <p>Due to the fact that, there are two different techniques of storing the two dimensional array into the memory, there are two different formulas to calculate the address of a random element of the 2D array. </p> <h3>By Row Major Order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = B. A. + (i * n + j) * size </pre> <p>where, B. A. is the base address or the address of the first element of the array a[0][0] . </p> <p> <strong>Example : </strong> </p> <pre> a[10...30, 55...75], base address of the array (BA) = 0, size of an element = 4 bytes . Find the location of a[15][68]. Address(a[15][68]) = 0 + ((15 - 10) x (68 - 55 + 1) + (68 - 55)) x 4 = (5 x 14 + 13) x 4 = 83 x 4 = 332 answer </pre> <h3>By Column major order </h3> <p>If array is declared by a[m][n] where m is the number of rows while n is the number of columns, then address of an element a[i][j] of the array stored in row major order is calculated as, </p> <pre> Address(a[i][j]) = ((j*m)+i)*Size + BA </pre> <p>where BA is the base address of the array. </p> <p> <strong>Example:</strong> </p> <pre> A [-5 ... +20][20 ... 70], BA = 1020, Size of element = 8 bytes. Find the location of a[0][30]. Address [A[0][30]) = ((30-20) x 24 + 5) x 8 + 1020 = 245 x 8 + 1020 = 2980 bytes </pre> <hr></3;i++)></pre></3;i++)></pre></3;i++)> kur B. A. ir bāzes adrese vai masīva pirmā elementa adrese a[0][0] .
Piemērs :
vārds
a[10...30, 55...75], base address of the array (BA) = 0, size of an element = 4 bytes . Find the location of a[15][68]. Address(a[15][68]) = 0 + ((15 - 10) x (68 - 55 + 1) + (68 - 55)) x 4 = (5 x 14 + 13) x 4 = 83 x 4 = 332 answer
Pēc kolonnas galvenā pasūtījuma
Ja masīvu deklarē ar a[m][n], kur m ir rindu skaits, bet n ir kolonnu skaits, tad masīva elementa a[i][j] adrese, kas saglabāta rindas galvenajā secībā, tiek aprēķināta kā ,
Address(a[i][j]) = ((j*m)+i)*Size + BA
kur BA ir masīva bāzes adrese.
Piemērs:
A [-5 ... +20][20 ... 70], BA = 1020, Size of element = 8 bytes. Find the location of a[0][30]. Address [A[0][30]) = ((30-20) x 24 + 5) x 8 + 1020 = 245 x 8 + 1020 = 2980 bytes